Chapter 11.4, Problem 85AYU

$\left(a\right)$

To determine

To calculate: The 2 by 3 and 3 by 2 matrices that represent the production of stainless steel and aluminum with different capacities.

Answer to Problem 85AYU

The 2 by 3 and 3 by 2 matrices that represent the production of stainless steel and aluminum on different days are,

  $\begin{array}{l}\begin{array}{cccccc}& & & 10\text{gal}& 5\text{gal}\begin{array}{cc}& \end{array}& \text{1gal}\end{array}\\ \begin{array}{c}\text{steel}\\ \text{aluminum}\end{array}\left[\begin{array}{ccccc}500& & 350& & 400\\ 700& & 500& & 850\end{array}\right]\\ \end{array}$ and $\begin{array}{l}\begin{array}{cccccc}& & & \text{Steel}& \text{Aluminum}\begin{array}{cc}& \end{array}& \end{array}\\ \begin{array}{c}\text{10gal}\\ \text{5gal}\\ \text{1gal}\end{array}\left[\begin{array}{ccccccc}& 500& & & 700& & \\ & 350& & & 500& & \\ & 400& & & 850& & \end{array}\right]\end{array}$ .

Explanation of Solution

Given information:

Number of stainless steel containers along with their capacities are, 500 with 10 gallon capacity, 350 with 5 gallon capacity and 400 with 1 gallon capacity.

Number of aluminum containers along with their capacities are, 700 with 10 gallon capacity, 500 with 5 gallon capacity and 850 with 1 gallon capacity.

Calculation:

Consider the provided information that the number of stainless steel containers along with their capacities are, 500 with 10 gallon capacity, 350 with 5 gallon capacity and 400 with 1 gallon capacity.

Number of aluminum containers along with their capacities are, 700 with 10 gallon capacity, 500 with 5 gallon capacity and 850 with 1 gallon capacity.

The above data can be represented as a $2\times 3$ where each row represent the number of steel and aluminum containers of different capacities.

  $\begin{array}{l}\begin{array}{cccccc}& & & 10\text{gal}& 5\text{gal}\begin{array}{cc}& \end{array}& \text{1gal}\end{array}\\ \begin{array}{c}\text{steel}\\ \text{aluminum}\end{array}\left[\begin{array}{ccccc}500& & 350& & 400\\ 700& & 500& & 850\end{array}\right]\\ \end{array}$

The above data can be represented as a $3\times 2$ where each row represent the number of steel and aluminum containers of different capacities.

  $\begin{array}{l}\begin{array}{cccccc}& & & \text{Steel}& \text{Aluminum}\begin{array}{cc}& \end{array}& \end{array}\\ \begin{array}{c}\text{10gal}\\ \text{5gal}\\ \text{1gal}\end{array}\left[\begin{array}{ccccccc}& 500& & & 700& & \\ & 350& & & 500& & \\ & 400& & & 850& & \end{array}\right]\end{array}$

$\left(b\right)$

To determine

To calculate: The matrix representation of amount of material used.

Answer to Problem 85AYU

The matrix representation of amount of material used is $\left[\begin{array}{c}15\\ 8\\ 3\end{array}\right]\begin{array}{c}\text{10gal}\\ \text{5gal}\\ \text{1gal}\end{array}$ .

Explanation of Solution

Given information:

The amount of material used in 10 gallon containers is 15 pounds, for 5 gallon containers is 8 pounds and for I gallon containers is 3 pounds.

It is provided that the amount of material used in 10 gallon containers is 15 pounds, for 5 gallon containers is 8 pounds and for I gallon containers is 3 pounds.

Construct a 3 by 1 column matrix to represent the amount of material used.

Therefore, amount of material used is

  $\left[\begin{array}{c}15\\ 8\\ 3\end{array}\right]\begin{array}{c}\text{10gal}\\ \text{5gal}\\ \text{1gal}\end{array}$

$\left(c\right)$

To determine

To calculate: The day’s usage of material.

Answer to Problem 85AYU

The usage of steel is 11500 pounds and aluminum is 17050 pounds.

Explanation of Solution

Given information:

The matrix representation of amount of material used is $\left[\begin{array}{c}15\\ 8\\ 3\end{array}\right]\begin{array}{c}\text{10gal}\\ \text{5gal}\\ \text{1gal}\end{array}$ .

The 2 by 3 matrix that represent the production of stainless steel and aluminum on different days is, $\begin{array}{l}\begin{array}{cccccc}& & & 10\text{gal}& 5\text{gal}\begin{array}{cc}& \end{array}& \text{1gal}\end{array}\\ \begin{array}{c}\text{steel}\\ \text{aluminum}\end{array}\left[\begin{array}{ccccc}500& & 350& & 400\\ 700& & 500& & 850\end{array}\right]\\ \end{array}$

It is provided that matrix representation of amount of material used is $\left[\begin{array}{c}15\\ 8\\ 3\end{array}\right]\begin{array}{c}\text{10gal}\\ \text{5gal}\\ \text{1gal}\end{array}$ .

The 2 by 3 matrix that represent the production of stainless steel and aluminum on different days is, $\begin{array}{l}\begin{array}{cccccc}& & & 10\text{gal}& 5\text{gal}\begin{array}{cc}& \end{array}& \text{1gal}\end{array}\\ \begin{array}{c}\text{steel}\\ \text{aluminum}\end{array}\left[\begin{array}{ccccc}500& & 350& & 400\\ 700& & 500& & 850\end{array}\right]\\ \end{array}$

Total day’s usage of material is

  $\begin{array}{c}\left[\begin{array}{ccccc}500& & 350& & 400\\ 700& & 500& & 850\end{array}\right]\left[\begin{array}{c}15\\ 8\\ 3\end{array}\right]=\left[\begin{array}{c}500\left(15\right)+350\left(8\right)+400\left(3\right)\\ 700\left(15\right)+500\left(8\right)+850\left(3\right)\end{array}\right]\\ =\left[\begin{array}{c}11500\\ 17050\end{array}\right]\end{array}$

Thus, day’s usage of steel is 11500 pounds and aluminum is 17050 pounds.

$\left(d\right)$

To determine

To calculate: The matrix representation of cost of steel and aluminum.

Answer to Problem 85AYU

The matrix representation of cost of steel and aluminum is $\begin{array}{l}\left[\begin{array}{cc}0.10& 0.05\end{array}\right]\\ \begin{array}{cc}\text{steel}& \text{aluminum}\end{array}\end{array}$ .

Explanation of Solution

Given information:

The cost of stainless steel is $\$0.10$ pounds and aluminum costs $\$0.05$ pounds.

It is provided that the cost of stainless steel is $\$0.10$ pounds and aluminum costs $\$0.05$ pounds.

Construct a 1 by 2 column matrix to represent the cost.

Therefore, cost of steel and aluminum is

  $\begin{array}{l}\left[\begin{array}{cc}0.10& 0.05\end{array}\right]\\ \begin{array}{cc}\text{steel}& \text{aluminum}\end{array}\end{array}$

$\left(d\right)$

To determine

To calculate: The total cost of day’s production.

Answer to Problem 85AYU

The total cost of day’s production is $\$2002.5$ .

Explanation of Solution

Given information:

The matrix representation of cost of steel and aluminum is $\begin{array}{l}\left[\begin{array}{cc}0.10& 0.05\end{array}\right]\\ \begin{array}{cc}\text{steel}& \text{aluminum}\end{array}\end{array}$ .

Total day’s usage of material is $\left[\begin{array}{c}11500\\ 17050\end{array}\right]$ .

It is provided that matrix representation of cost of steel and aluminum is $\begin{array}{l}\left[\begin{array}{cc}0.10& 0.05\end{array}\right]\\ \begin{array}{cc}\text{steel}& \text{aluminum}\end{array}\end{array}$ .

Total day’s usage of material is $\left[\begin{array}{c}11500\\ 17050\end{array}\right]$ .

Total cost of day’s production is

  $\begin{array}{c}\left[\begin{array}{cc}0.10& 0.05\end{array}\right]\left[\begin{array}{c}11500\\ 17050\end{array}\right]=0.10\left(11500\right)+0.05\left(17050\right)\\ =2002.5\end{array}$

Thus, total cost of day’s production is $\$2002.5$ .